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The sum of perimeter of a circle & square is K.where K is some constant.Prove that the sum of their areas is least when the side of square is double

the radius of circle.



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Let side of square = a
radius of circle = r
perimeter of square = 4a
perimeter of circle = 2πr
sum = 4a+2πr = K
⇒4a = K-2πr
⇒a = (K-2πr)/4

area of square = a² = ((K-2πr)/4)² = (K-2πr)²/16
area of circle = πr²
sum of areas, S = πr² + (K-2πr)²/16

For minimum area,  \frac{dS}{dr} =0

 \frac{d}{dr} ( \pi  r^{2}+  \frac{ (K-2 \pi r)^{2} }{16} )=0

 \frac{d}{dr} ( \pi  r^{2})+   \frac{d}{dr}( \frac{ (K-2 \pi r)^{2} }{16} )=0

2 \pi r+ \frac{1}{16}(8  \pi ^{2} r-4 \pi K)=0

2 \pi r+ \frac{1}{16}(8  \pi ^{2} r-4 \pi (4a+2 \pi r))=0

2 \pi r+ \frac{1}{16}(8  \pi ^{2} r-16 \pi a-8 \pi^{2}  r))=0

2 \pi r+ \frac{1}{16}(-16 \pi a)=0

2 \pi r- \pi a=0

2r=a  (proved)

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P(circle) =2pie r
  2pie r+4a=k
  pie r2 +a2 
 (given that) 2r=a
 assuming the value of r=1
then, 2*22/7*1+8(1)=100/7 (sum of peri. )
so here area is minimum
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